Shell: Getting Arguments

kalyanofb · 03-01-2007, 02:03 AM

getDetails()
{
for i in $1; do

done
}

getDetails outfn arg1 arg2 arg3

The above is script model. I want to store some information to argument1 while processing for loop. How i will do this.

Help me plz.

puffinman · 03-01-2007, 02:31 AM

Not quite sure exactly what you mean, but is it something along these lines?

Code:

getDetails() {
  OUTFN=$1;
  shift;
for arg in $*;
  do
  #do whatever you want to do here...
  echo $arg >> $OUTFN
done
}

kalyanofb · 03-01-2007, 02:36 AM

How i will skip the first argument inside the loop

puffinman · 03-01-2007, 03:25 AM

That's what the shift does...

cfaj · 03-01-2007, 11:20 AM

Quote:

Originally Posted by puffinman

Not quite sure exactly what you mean, but is it something along these lines?

Code:

getDetails() {
  OUTFN=$1;
  shift;
for arg in $*;
  do
  #do whatever you want to do here...
  echo $arg >> $OUTFN
done
}

Use "$@" not $*, printf rather than echo, and quote your variables:

Code:

getDetails() {
  OUTFN=$1;
  shift;
  for arg in "$@"
  do
    #do whatever you want to do here...
    printf "%s\n" "$arg"
  done > $OUTFN
}